Problem: What is the radius of the circle inscribed in triangle $ABC$ if $AB = 22, AC=12,$ and $BC=14$? Express your answer in simplest radical form.
Let $r$ be the radius of the inscribed circle. Let $s$ be the semiperimeter of the triangle, that is, $s=\frac{AB+AC+BC}{2}=24$. Let $K$ denote the area of $\triangle ABC$.

Heron's formula tells us that \begin{align*}
K &= \sqrt{s(s-AB)(s-AC)(s-BC)} \\
&= \sqrt{24\cdot 2\cdot 12\cdot 10} \\
&= \sqrt{24^2\cdot 10} \\
&= 24\sqrt{10}.
\end{align*}The area of a triangle is equal to its semiperimeter multiplied by the radius of its inscribed circle ($K=rs$), so we have $$24\sqrt{10} = r\cdot 24,$$which yields the radius $r=\boxed{\sqrt{10}}$.